A number of clever examples have been given of unbounded connected subsets of the Euclidean plane containing no infinite bounded subsets that are connected. None of those that I have seen are completely metrizable. Does anybody know if such can exist or if their existence can be ruled out by some theorem? I know that no such completely metrizable sets can exist if they are the graphs of functions of the form y=f(x) in the Cartesian plane. But does this prohibition extend to all unbounded connected planar sets?

2$\begingroup$ will you clarify better the example you are seeking? "Containing no bounded subsets that are connected" is not clear to me (an unbounded set certainly contains a singleton, which is a connected subset) $\endgroup$ – Pietro Majer Aug 24 '10 at 19:54

1$\begingroup$ I agree with Pietro: "containing no bounded subsets that are connected" is definitely wrong; a single point is a bounded, connected subset! Sorry, I can't guess your intended meaning. $\endgroup$ – Zen Harper Aug 25 '10 at 3:04

$\begingroup$ All of the connected subsets I am talking about are supposed to be infinite, whether they are bounded or unbounded. Any connected subset of a metric space that contains more than one point is infinite. Perhaps I should have stated this in my question but I thought it would be clear. Allowing singletons to be bounded connected sets would make the problem trivial or nonsensical. $\endgroup$ – Garabed Gulbenkian Aug 25 '10 at 17:54

2$\begingroup$ Yes, of course you should better state it clearly, e.g. saying "no infinite bounded subsets that are connected". The fact that the problem becomes trivial or nonsensical otherwise, is not sufficient to make it clear what you mean, as singletons definitely are bounded connected sets. $\endgroup$ – Pietro Majer Aug 26 '10 at 9:08
The following gives a partial answer: no such unbounded connected set may exist with the further assumption that it is closed. Actually, the argument generalizes for any locally compact metric space. I'm not completely sure that a much simpler or even trivial proof may exist, though.
Let $\Gamma$ be a closed unbounded connected subset of the plane. Let $x\in\Gamma$ and let $B:=B(x,r)$ be an open ball around $x$. I claim that the connected component of $x$ in $\Gamma\cap \bar{B}$ meets $\partial B$, which shows that $\Gamma$ does contain nontrivial bounded connected subsets.
For any $\epsilon>0$, consider the $\epsilon$neighborhood of $\Gamma,$ that is $\Gamma_\epsilon:=\cup_{y\in\Gamma}B(y,\epsilon).$ It is an open unbounded connected subset of the plane. Let $U_\epsilon$ be the connected component of $x$ in $\Gamma_\epsilon\cap B$. Since the latter is locally connected, $U_\epsilon$ is both an open and closed subset of it in the relative topology. It is therefore an open subset of $\Gamma_\epsilon$; however it is not closed in it, because $\Gamma_\epsilon$ is connected. Therefore $\bar U_\epsilon$ is a closed connected set that meets $\partial B, $ and of course contains $x$. Since the set of all connected closed subsets of a compact metric space is compact in the Hausdorff distance, taking a limit as $\epsilon\to0$ we get a bounded connected subset of $\Gamma$ connecting $x$ with $\partial B$ (this also passes to the limit).
Rmk One could state the above in terms of the onepoint compactification of $\Gamma$, and more generally for compact connected metric spaces. The trick of approximating a metric space with a locally connected metric space is made possible via the Kuratowski embedding (one defines $X_\epsilon$ as an $\epsilon$ nbd of $X$ in the embedding).
PS: Of course the same affirmative conclusion holds, even more directely, if $\Gamma$ is assumed to be open, which is another case included in the original assumption of completely metrizable.

1$\begingroup$ "Since the set of all connected closed subsets of a compact metric space is compact in the Hausdorff distance": all you need is a more elementary fact that the intersection of a nested family of continua (connected compact metric spaces) is a continuum. $\endgroup$ – Victor Protsak Aug 27 '10 at 9:01

$\begingroup$ You are right! I didn't realize that those set are nested. However I like so much recalling the glorious name of Felix Hausdorff... $\endgroup$ – Pietro Majer Aug 27 '10 at 20:12

$\begingroup$ Thanks alot for your impressively reasoned responses which rule out point sets that are closed and locally compact. Indeed, all the specific examples I have seen in the literature were of unbounded connected subsets of the plane which were not closed. These sets contained no bounded connected subsets except sigletons. However none of these examples was completely metrizable. Perhaps one might have better luck with subsets of Hilbert space which can be closed and are therefore completely metrizablewithout necessarily being locally compact. $\endgroup$ – Garabed Gulbenkian Aug 28 '10 at 15:20